Worked Example 3
3.8 Cumulative Frequency
Cumulative frequencies are useful if more detailed information is required about a set of data. In particular, they can be used to find the median and inter-quartile range.
The inter-quartile range contains the middle 50% of the sample and is a measure of how spread out the data are. This is illustrated in Worked Example 2.
Worked Example 1
The heights of 112 children aged 5 to 11 are recorded in this table. Draw up a cumulative frequency table and then draw a cumulative frequency polygon.
Solution
The table below shows how to calculate the cumulative frequencies.
Height (cm) Frequency Cumulative Frequency
90 < ≤h 100 5 5
A graph can then be plotted using points as shown below.
Height (cm) Frequency means that there are 27 values in the interval
90< ≤h 110. In the same way, 57 values are at most 120 cm, so we plot 57 against 120 cm on the graph.
Note
If the table in Worked Example 1 gives the frequencies in thousands for each height group in a particular town, then we represent this by a cumulative frequency curve, showing the distribution for that whole population.
Worked Example 2
The cumulative frequency graph below gives the results of 120 students on a test.
Height (cm)
0 20 40 60 80 100 120
90 100 110 120 130 140 150
(90,0) (100,5)
(110,27)
(120,57)
(130,88)
(140,106) (150,112)
Cumulative Frequency
0 20 40 60 80 100 120
20 40 60 80 100
0
Use the graph to find:
(a) the median score, (b) the inter-quartile range,
(c) the mark which was attained by only 10% of the students, (d) the number of students who scored more than 75 on the test.
Solution
(a) Since 12 of 120 is 60, the median can be found by starting at 60 on the vertical scale, moving horizontally to the graph line and then moving vertically down to meet the horizontal scale.
In this case the median is 53.
(b) To find out the inter-quartile range, we must consider the middle 50% of the students.
To find the lower quartile, start at 14 of 120, which is 30.
This gives
Lower Quartile = 43
To find the upper quartile, start at 34 of 120, which is 90.
Lower quartile = 43
Upper quartile = 67
Cumulative Frequency
(c)
Here the mark which was attained by the top 10% is required.
10% of120 = 12
so start at 108 on the cumulative frequency scale.
This gives a mark of 79, so the top 10% scored 79 or higher on this test.
(d) To find the number of students who scored more than 75, start at 75 on the horizontal axis.
This gives a cumulative frequency of 103.
So the number of students with a score greater than 75 is
120−103 = 17
To find the median in a small sample we should use the n+1
2 rule (see Section 3.3 on Mean, Median and Mode). However, for this example the difference between using the
6012th value and the 60th value is minimal and so the method shown here is appropriate for this and all other large samples. (This also applies to the finding of quartiles, percentiles, deciles, etc.)
Exercises
1. Make a cumulative frequency table for each set of data given below. Then draw a cumulative frequency graph and use it to find the median and inter-quartile range.
(a) John weighed each apple in a large box. His results are given in this table.
Weight of
apple (g) 60< w≤ 80 80 <w ≤100 100< w≤120 120< w≤140 140< w≤160
Frequency 4 28 33 27 8
(b) Pasi asked the students in his class how far they travelled to school each day.
His results are given below.
Distance (km) 0< ≤d 1 1< ≤d 2 2< ≤d 3 3< ≤d 4 4 < ≤d 5 5< ≤d 6
Frequency 5 12 5 6 5 3
(c) A P.E. teacher recorded the distances children could reach in the long jump event. His records are summarised in the table below.
Length of jump (m) 1< ≤d 2 2 < ≤d 3 3 < ≤d 4 4< ≤d 5 5< ≤d 6
Frequency 5 12 5 6 5
2. A farmer grows a type of wheat in two different fields. He takes a sample of 50 heads of corn from each field at random and weighs the grains he obtains.
Mass of grain (g) 0< ≤m 5 5< ≤m 10 10< ≤m 15 15< ≤m 20 20< ≤m 25 25< ≤m 30
Frequency Field A 3 8 22 10 4 3
Frequency Field B 0 11 34 4 1 0
(a) Draw cumulative frequency graphs for each field.
(b) Find the median and inter-quartile range for each field.
(c) Comment on your results.
3. A consumer group tests two types of batteries using a personal stereo.
(a) Use cumulative frequency graphs to find the median and inter-quartile range for each type of battery.
(b) Which type of battery would you recommend and why?
4. The table below shows how the height of girls of a certain age vary. The data was gathered using a large-scale survey.
Height (cm) 50< ≤h 55 55< ≤h 60 60< ≤h 65 65< ≤h 70 70< ≤h 75 75< ≤h 80 80< ≤h 85
Frequency 100 300 2400 1300 700 150 50
A doctor wishes to be able to classify children as:
Category Percentage of Population
Very Tall 5%
Tall 15%
Normal 60%
Short 15%
Very short 5%
Use a cumulative frequency graph to find the heights of children in each category.
5. The manager of a double glazing company employs 30 salesmen. Each year he awards bonuses to his salesmen.
Bonus Awarded to
£500 Best 10% of salesmen £250 Middle 70% of salesmen £ 50 Bottom 20% of salesmen
The sales made during 1995 and 1996 are shown in the table below.
Value of sales
(£1000) 0< ≤V 100 100< ≤V 200 200< ≤V 300 300< ≤V 400 400< ≤V 500
Frequency 1995 2 8 18 2 0
Frequency 1996 0 2 15 10 3
Use cumulative frequency graphs to find the values of sales needed to obtain each bonus in the years 1995 and 1996.
6. Laura and Joy played 40 games of golf together. The table below shows Laura's scores.
Scores (x) 70< ≤x 80 80 < ≤x 90 90< ≤x 100 100< ≤x 110 110 < ≤x 120
Frequency 1 4 15 17 3
(a) On a grid similar to the one below, draw a cumulative frequency diagram to show Laura's scores.
(b) Making your method clear, use your graph to find (i) Laura's median score,
(ii) the inter-quartile range of her scores.
(c) Joy's median score was 103. The inter-quartile range of her scores was 6.
(i) Who was the more consistent player?
Give a reason for your choice.
(ii) The winner of a game of golf is the one with the lowest score.
Who was the better player? Give a reason for your choice.
(NEAB)
Cumulative Frequency
Score
60 70 80 90 100 110 120
10 20 30 40
qy
7. A sample of 80 electric light bulbs was taken. The lifetime of each light bulb was recorded. The results are shown below.
Lifetime (hours) 800– 900– 1000– 1100– 1200– 1300– 1400–
Frequency 4 13 17 22 20 4 0
Cumulative Frequency 4 17
(a) Copy and complete the table of values for the cumulative frequency.
(b) Draw the cumulative frequency curve, using a grid as shown.
(c) Use your graph to estimate the number of light bulbs which lasted more than 1030 hours.
(d) Use your graph to estimate the inter-quartile range of the lifetimes of the light bulbs.
(e) A second sample of 80 light bulbs has the same median lifetime as the first sample. Its inter-quartile range is 90 hours. What does this tell you about the difference between the two samples?
(SEG)
800 900 1000 1100 1200 1300 1400 1500
20 10 0 30 40 50 60 70 80 90
qyCumulative Frequency
Lifetime (hours)
8. The numbers of journeys made by a group of people using public transport in one month are summarised in the table.
Number of journeys 0–10 11–20 21–30 31–40 41–50 51–60 61–70
Number of people 4 7 8 6 3 4 0
(a) Copy and complete the cumulative frequency table below.
Number of journeys ≤10 ≤20 ≤30 ≤40 ≤50 ≤60 ≤70 Cumulative frequency
(b) (i) Draw the cumulative frequency graph, using a grid as below.
(ii) Use your graph to estimate the median number of journeys.
(iii) Use your graph to estimate the number of people who made more than 44 journeys in the month.
(c) The numbers of journeys made using public transport in one month, by another group of people, are shown in the graph.
Number of journeys
10 20 30 40 50 60 70
10
0 20 30 40
qyCumulative Frequency
10 20 30 40
Cumulative Frequency
9. The lengths of a number of nails were measured to the nearest 0.01 cm, and the following frequency distribution was obtained.
Length of nail Number of nails Cumulative Frequency (x cm)
0 98. ≤ <x 1 00. 2 1 00. ≤ <x 1 02. 4 1 02. ≤ <x 1 04. 10 1 04. ≤ <x 1 06. 24 1 06. ≤ <x 1 08. 32 1 08. ≤ <x 1 10. 17 1 10. ≤ <x 1 12. 7 1 12. ≤ <x 1 14. 4
(a) Complete the cumulative frequency column.
(b) Draw a cumulative frequency diagram on a grid similar to the one below.
Use your graph to estimate
(i) the median length of the nails (ii) the inter-quartile range.
(NEAB)
0 0.98 1.00 1.02 1.04 1.06 1.08 1.10 1.12 1.14
20 40 60 80 100
Length of nail (cm)
10. A wedding was attended by 120 guests. The distance, d miles, that each guest travelled was recorded in the frequency table below.
Distance
(d miles) 0< ≤d 10 10< ≤d 20 20< ≤d 30 30< ≤d 50 50< ≤d 100 100< ≤d 140 Number of
guests 26 38 20 20 12 4
(a) Using the mid-interval values, calculate an estimate of the mean distance travelled.
(b) (i) Copy and complete the cumulative frequency table below.
Distance
(d miles) d≤ 10 d ≤ 20 d ≤ 30 d ≤ 50 d ≤ 100 d ≤ 140 Number of
guests 120
(ii) On a grid similar to that shown below, draw a cumulative frequency curve to represent the information in the table.
(c) (i) Use the cumulative frequency curve to estimate the median distance
0 20 40 60 80 100 120 140
20 40 60 80 100 120
Distance travelled (miles) Cumulative
Frequency